Prove that the straight line x + y = 2 + 2 touches the circle x 2 + y 2 - 2 x - 2 y + 1 = 0 . Find the point of contact.

Given that, the equation of the circle: x 2 + y 2 - 2 x - 2 y + 1 = 0 Here, centre = 1 , 1 and radius 1 + 1 - 1 = 1 So, the distance of the centre 1 , 1 from the line x + y = 2 + 2 is d = 1 + 1 - 2 - 2 2 = 1 Thus, clearly here Radius = The distance of the centre 1 , 1 from the line x + y = 2 + 2 Hence, the given line x + y = 2 + 2 touches the circle x 2 + y 2 - 2 x - 2 y + 1 = 0 Now, differentiating the circle both sides w.r.t x , we get: 2 x + 2 y × d y d x - 2 - 2 × d y d x = 0 ⇒ d y d x 2 y - 2 = 2 - 2 x ⇒ d y d x = 2 - 2 x 2 y - 2 Now, the slope of the tangent at x 1 , y 1 is d y d x x 1 , y 1 = ( 2 - 2 x 1 ) ( 2 y 1 - 2 ) Hence, the equation of the tangent at x 1 , y 1 with slope 2 - 2 x 1 2 y 1 - 2 is y - y 1 = 2 - 2 x 1 2 y 1 - 2 x - x 1 [Using y − y 1 = d y d x ( x 1 , y 1 ) x − x 1 ] ⇒ 2 y 1 - 2 y - y 1 = 2 - 2 x 1 x - x 1 ⇒ 2 y y 1 - 2 y 2 1 - 2 y + 2 y 1 = 2 x - 2 x 1 - 2 x x 1 + 2 x 2 1 ⇒ 2 y y 1 - 1 - 2 y 2 1 + 2 y 1 = 2 x 1 - x 1 - 2 x 1 + 2 x 2 1 ⇒ 2 x 1 - x 1 - 2 y y 1 - 1 - 2 x 1 + 2 x 2 1 + 2 y 2 1 - 2 y 1 = 0 ⇒ x 1 - x 1 - y y 1 - 1 - x 1 + x 2 1 + y 2 1 - y 1 = 0 Thus, we have the identical equations, x + y = 2 + 2 and x 1 - x 1 - y y 1 - 1 - x 1 + x 2 + 1 y 1 2 - y 1 = 0 [both being the equations of same tangent] So, 1 1 - x 1 = 1 - y 1 - 1 = 2 + 2 x 1 - x 1 2 - y 1 2 + y 1 ⇒ 1 1 - x 1 = 2 + 2 x 1 - x 1 2 - y 1 2 + y 1 → 1 ; 1 - y 1 - 1 = 2 + 2 x 1 - x 1 2 - y 1 2 + y 1 → 2 ; 1 1 - x 1 = 1 - y 1 - 1 → 3 Now, from third equation, we have: x 1 = y 1 Substituting x 1 = y 1 in 1 and 2 , we get: x 1 = 1 + 1 2 and y 1 = 1 + 1 2 Hence, the point of contact is x 1 , y 1 = 1 + 1 2 , 1 + 1 2 .

Prove that the tangent to the curve x 3 + y 3 = 3 a x y at 3 a 2 , 3 a 2 makes obtuse angle with the positive direction of the x - axis.

Given that, the equation of curve x 3 + y 3 = 3 a x y Differentiating the curve both sides w.r.t x , we get: 3 x 2 + 3 y 2 d y d x = 3 a x d y d x + y ⇒ x 2 + y 2 d y d x = a x d y d x + a y ⇒ y 2 d y d x - a x d y d x = a y - x 2 ⇒ d y d x y 2 - a x = a y - x 2 ⇒ d y d x = a y - x 2 y 2 - a x Now, the slope of the tangent at 3 a 2 , 3 a 2 is d y d x 3 a 2 , 3 a 2 = a × 3 a 2 - 3 a 2 2 3 a 2 2 - a × 3 a 2 = 3 a 2 2 - 3 a 2 2 3 a 2 2 - 3 a 2 2 = - 1 Now, the slope of the tangent m = d y d x 3 a 2 , 3 a 2 = tan θ = - 1 ⇒ tan θ = tan 3 π 4 ⇒ θ = 3 π 4 or 135 0 So, clearly here θ lies between 90 0 and 180 0 i.e., it is an obtuse angle. Thus, the tangent will make an obtuse angle with the positive direction of x - axis.

E M B I B E

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Calculus - C. Applications of Calculus>Chapter 18 - Tangent and Normal>EXERCISE

Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Author:Parthasarathi Mukhopadhyay & Manabendra Nath Mukherjee

Parthasarathi Mukhopadhyay Mathematics Solutions for Exercise - Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Attempt the practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. RUDIMENTS of MATHEMATICS For Class 12 of +2 Level solutions are prepared by Experienced Embibe Experts.

Questions from Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE with Hints & Solutions

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 18 - Tangent and Normal>EXERCISE>Q 11

Find the co-ordinates of the point at which the normal to the parabola $y^{2} = x$ perpendicular to the line $x + y = 1$ .

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 18 - Tangent and Normal>EXERCISE>Q 12

A plane curve is given by the parametric equation, $x = 3 \cos θ - \cos^{3} θ, y = 3 \sin θ - \sin^{3} θ$ . Find the equation of the normal at any point on the curve and show that, at the point $θ = \frac{π}{4},$ the normal passes through the origin.

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 18 - Tangent and Normal>EXERCISE>Q 13

Prove that the straight line $x + y = 2 + \sqrt{2}$ touches the circle $x^{2} + y^{2} - 2 x - 2 y + 1 = 0$ . Find the point of contact.

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 18 - Tangent and Normal>EXERCISE>Q 14

Find the co-ordinates of the point on the curve $y = 1 + 2 x - 3 x^{2}$ at which the tangent makes an angle $45 °$ with the positive direction of the $x -$ axis.

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 18 - Tangent and Normal>EXERCISE>Q 15

Prove that the tangent to the curve $x^{3} + y^{3} = 3 a x y$ at $(\frac{3 a}{2}, \frac{3 a}{2})$ makes obtuse angle with the positive direction of the $x -$ axis.